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aalyn [17]
3 years ago
6

Which table shows a proportional relationship between x and y?​

Mathematics
2 answers:
nekit [7.7K]3 years ago
6 0

Answer:D

Step-by-step explanation:

This is D because your dividing by 5 or each one and thats how you get Your output number

Molodets [167]3 years ago
4 0

Answer:

D

Step-by-step explanation:

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Hi need this done fast, please
Blizzard [7]
Length:
60% of 6 is 3.6

3.6 x 6 = 21.6


Width:
150% of 5 is 7.5

7.5 x 5 = 37.5

So her mothers garden should be 21.6 feet in length, and 37.5 feet in width.

5 0
3 years ago
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Which expression is equivalnt (-18) - 64n
Naily [24]

Answer:

Your correct answer is option C. -2 (9 + 32n)

Step-by-step explanation:

Solution:

Options are not given in the question

Given First degree expression:

-18-64n

= (-2)×9 + (-2)×32n

Take (-2) common, we get

= (-2)( 9+32n)

Therefore you get your answer of option C. -2(9+32n)

8 0
3 years ago
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The gcf of 96x^2 and 88x is
PolarNik [594]
The GCF of 96x² and 88x is 8x.
5 0
3 years ago
Jack works as a waiter and is keeping track of the tips he earns daily. about how much does jack have to earn in tips on sunday
OLga [1]

A mean is an arithmetic average of a set of observations. The amount that Jack has to earn in tips on Sunday if he wants to average $19 a day is $25.

<h3>What is Mean?</h3>

A mean is an arithmetic average of a set of observations. it is given by the formula,

\rm Mean=\dfrac{\text{Sum of all obervation}}{\text{Number of observation}}

As it is given that Jack makes $12 on Monday, $14 on Tuesday, $18 on Wednesday, $16 on Thursday, $26 on Friday and $22 on Saturday. And we need to know how much he should earn on Sunday, so the average is $19. Therefore, we can write,

\rm Mean=\dfrac{\text{Sum of all obervation}}{\text{Number of observation}}

19 = \dfrac{12+14+18+16+26+22+x}{7}\\\\133 = 108 + x\\\\133-108=x\\\\x = 25

Hence, the amount that Jack has to earn in tips on Sunday if he wants to average $19 a day is $25.

Learn more about Mean:

brainly.com/question/16967035

5 0
2 years ago
A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the
Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3.6, \sigma = 0.4

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

X = 3.6

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.6 - 3.6}{0.4}

Z = 0

Z = 0 has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

Z = \frac{X - \mu}{\sigma}

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 4.2

Z = \frac{X - \mu}{\sigma}

Z = \frac{4.2 - 3.6}{0.4}

Z = 1.5

Z = 1.5 has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

Z = \frac{X - \mu}{\sigma}

Z = \frac{5 - 3.6}{0.4}

Z = 3.5

Z = 3.5 has a pvalue of 0.9998

X = 3

Z = \frac{X - \mu}{\sigma}

Z = \frac{3 - 3.6}{0.4}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

Z = \frac{X - \mu}{\sigma}

1.75 = \frac{X - 3.6}{0.4}

X - 3.6 = 0.4*1.75

X = 4.3

They last at least 4.3 minutes

7 0
3 years ago
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